1 Answer
1

You may find an asymptotic formula for $a_n$ by improving the accuracy in an adaptive manner.

Step 1. Since
$$0 < a_{n+1} = \log (1+a_n) < a_n,$$
it is a monotone decreasing sequence which is bounded. Thus it must converge to some limit, say $\alpha$. Then $\alpha = \log(1 + \alpha)$, which is true precisely when $\alpha = 0$. Therefore it follows that
$$a_n = o(1). \tag{1}$$

Before going to the next step, we make a simple observation: it is easy to observe that the function
$$\frac{x}{\log(1+x)}$$
is of class $C^3$. In particular, whenever $|x| \leq \frac{1}{2}$, we have
$$ \frac{x}{\log (1+x)} = 1+\frac{x}{2}-\frac{x^2}{12}+O(x^3). $$
This can be rephrased as
$$ \frac{1}{\log(1+x)} = \frac{1}{x}+\frac{1}{2}-\frac{x}{12}+O(x^2). \tag{2}$$
Here, we note that the bound, say $C > 0$, for the Big-Oh notation does not depend on $x$ whenever $|x| \leq \frac{1}{2}$.